Solving Derivation Problem: \sum_{j=\kappa}^{\inf} \varphi^{2j-\kappa}\sigma^2

[roadworx]

Hi,

I have $$\sum_{j=\kappa}^{\inf} \varphi^{2j-\kappa}\sigma^2$$

I know the answer is $$= \frac{\sigma^2}{1-\sigma^2} \varphi^\kappa$$

Can someone explain the mathematics involved in this derivation?

Thanks.

 

[slider142]

Does $\sigma$ have an exponent related to j? Otherwise, the sum is a straightforward geometric series, once you reparameterize the index of summation. Note that
$$\sum_{j=\kappa}^\infty \varphi^{2j-\kappa} = \varphi^\kappa + \varphi^{\kappa + 2} + \varphi^{\kappa + 6} + \cdots = \varphi^{\kappa}\sum_{n=0}^\infty \varphi^{2n} = \varphi^{\kappa}\sum_{n=0}^\infty (\varphi^2)^n$$
Do you know the sum of a geometric series?

 

[roadworx]

Does $\sigma$ have an exponent related to j? Otherwise, the sum is a straightforward geometric series, once you reparameterize the index of summation. Note that
$$\sum_{j=\kappa}^\infty \varphi^{2j-\kappa} = \varphi^\kappa + \varphi^{\kappa + 2} + \varphi^{\kappa + 6} + \cdots = \varphi^{\kappa}\sum_{n=0}^\infty \varphi^{2n} = \varphi^{\kappa}\sum_{n=0}^\infty (\varphi^2)^n$$
Do you know the sum of a geometric series?

Thanks for that. I have a question on your reparameterization. Isn't $$\varphi^{\kappa}$$ to the negative index, and therefore shouldn't it be $$\frac{1}{\varphi^{\kappa}}$$ when you write out the series? Or did I miss something?

 

[slider142]

$\kappa$ is the same as that in the original sum. When j starts at $\kappa$, we get 2K - K = K, 2(K + 1) - K = K + 2, 2(K + 2) - K = K + 4, and so on.